Normal modes of vibration from the total energy

In summary: The correct Lagrangian is $$ L = \dfrac{1}{2}m(1+4a^2x^2)\dot{x}^2 + \dfrac{1}{2}M(\dot{x}^2+\dot{y}^2) -mgax^2 - Mgy + \dfrac{1}{2}k((ax^2-y-z)-l_0)^2\ , $$ and that makes the correct equations of motion be $$ \left\{\begin{array}{l} (m+M)\ddot{x} + 2a\left[mg-k(ax^2-z+l_0)\right]x = 0 \\ M\ddot{y
  • #1
Jaime_mc2
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A mass ##m## is restricted to move in the parabola ##y=ax^2##, with ##a>0##. Another mass ##M## is hanging from this first mass using a spring with constant ##k## and natural lenghth ##l_0##. The spring is restricted to be in vertical position always. The coordinates for the system are ##x## (horizontal coordinate of mass ##m##) and ##y## (vertical coordinate of mass ##M##).

With this system in mind, I was told to look for the equilibrium points using Lagrangian formulation, then approximate the kinetic and potential energy for small oscillations around that point, and finally, obtain the small oscillation frequency and normal modes of vibration from the total energy.

The first part was quite easy. I found the Lagrangian $$ L = \dfrac{1}{2}m(1+4a^2x^2)\dot{x}^2 + \dfrac{1}{2}M(\dot{x}^2+\dot{y}^2) -mgax^2 - Mgy - \dfrac{1}{2}k(ax^2-y-l_0)^2\ , $$ and applying the Euler-Lagrange equations I found the equations of motion, from where I could calculate the equilibrium point ##(x_0,y_0) = (0,y_0)##.

Now it is where the problems start to appear. By using a Taylor expansion around the equilibrium point, I found the approximations for the kinetic and potential energy, which are $$ T \approx \dfrac{1}{2}(m+M)\dot{x}^2 + \dfrac{1}{2}M\dot{y}^2 \quad\text{y}\quad V \approx a\left[mg-k(y_0+l_0)\right]x^2 + \dfrac{1}{2}k(y-y_0)^2\ , $$ where I removed the constant term ##V(0,y_0)## for simplicity.

I don't understand how I am suppose to get the normal modes of oscillation from the total energy. I thought about applying energy conservation but, apparently, this is only recommended when the system is one-dimensional. Since the Lagrangian does not depend on time, neither the coordinates and potential, I assumed that what they wanted is to associate the total energy with the Hamiltonian, so I computed it: $$ H \approx \dfrac{p_x^2}{2(m+M)} + \dfrac{p_y^2}{2M} + a\left[mg-k(y_0+l_0)\right]x^2 + \dfrac{1}{2}k(y-y_0)^2 $$ Applying the cannonical equations, what I get is the system $$ \left\{\begin{array}{l} (m+M)\ddot{x} + 2a\left[mg-k(y_0+l_0)\right]x = 0 \\ M\ddot{y} + k(y-y_0) = 0 \end{array}\right.\ , $$ which is not coupled, and ##x## and ##y## seem to oscillate in different independent frequencies.

How am I supposed to get normal modes from a system where the variables oscillate independently? Also, is there any other way in which I can use the total energy to solve the problem?
 
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  • #2
Jaime_mc2 said:
How am I supposed to get normal modes from a system where the variables oscillate independently?
I can't see anything wrong with your development so, if correct, aren't the modes are already normal?

Okay, what happen to the constraint ##y=ax^2##?
 
  • #3
Okay, I think ##y## is the contact point with the parabola. One must also account for the length of the spring with an additional variable, say ##z##. ##x## determines ##y## through the constraint.
 
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  • #4
Paul Colby said:
Okay, I think ##y## is the contact point with the parabola. One must also account for the length of the spring with an additional variable, say ##z##. ##x## determines ##y## through the constraint.

I think you are right, the mistake is even in the problem statement:
Jaime_mc2 said:
A mass ##m## is restricted to move in the parabola ##y=ax^2##, with ##a>0##. Another mass ##M## is hanging from this first mass using a spring with constant ##k## and natural lenghth ##l_0##. The spring is restricted to be in vertical position always. The coordinates for the system are ##x## (horizontal coordinate of mass ##m##) and ##y## (vertical coordinate of mass ##M##).

when really the vertical coordinate of ##M## is not ##y##, but rather ##y - z = ax^2 - z##. That changes a few parts of the Lagrangian!
 
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Related to Normal modes of vibration from the total energy

1. What are normal modes of vibration?

Normal modes of vibration refer to the different ways in which a system can oscillate or vibrate. These modes are determined by the system's energy and geometry, and each mode has a specific frequency and amplitude.

2. How are normal modes of vibration related to total energy?

The total energy of a system is directly related to its normal modes of vibration. The more modes a system has, the more energy it can store. Additionally, the total energy of a system is distributed among its normal modes, with each mode having a specific amount of energy.

3. How are normal modes of vibration calculated?

The calculation of normal modes of vibration involves solving a mathematical equation known as the wave equation. This equation takes into account the system's mass, stiffness, and damping properties to determine the different modes of vibration.

4. What is the significance of normal modes of vibration?

Normal modes of vibration are important in understanding the behavior of various systems, such as molecules, atoms, and mechanical structures. They can also be used to predict the response of a system to external forces and to design systems with desired vibrational characteristics.

5. How do normal modes of vibration affect the stability of a system?

The normal modes of vibration of a system can affect its stability by determining how it responds to external forces. If a system is excited at one of its natural frequencies (corresponding to a normal mode), it can experience resonance and potentially become unstable. Therefore, understanding a system's normal modes is crucial in ensuring its stability.

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